IB Chemistry Free-Radical Substitution: FAQs

Free-radical substitution is the reaction where a halogen — usually chlorine or bromine — replaces a hydrogen atom on an alkane, kicked off by UV light splitting the halogen molecule into two reactive radicals. It's core Reactivity content in the current DP Chemistry guide, first examined 2025, taught at both SL and HL.

Because the propagation step isn't selective — a chlorine radical can abstract any hydrogen on the chain, and the halogenoalkane formed can react again before the chain ever terminates. The result is a mixture: chloromethane, dichloromethane, trichloromethane and tetrachloromethane, plus traces of ethane from radical-radical collisions.

Homolytic fission splits a covalent bond so each atom keeps one electron, forming two radicals — shown with single-barbed (fish-hook) arrows. Heterolytic fission gives both bonding electrons to one atom, forming a cation and an anion, shown with full double-barbed arrows. Radical substitution relies entirely on homolytic fission.

It appears on Paper 1 as multiple-choice questions — naming products, spotting initiation, propagation or termination steps — and on Paper 2 as extended-response questions asking you to draw the full mechanism with curly arrows. Since the current DP Chemistry guide (first exams 2025) removed Paper 3, mechanism questions can also sit alongside data-based questions on Paper 2.

Draw it in four stages: initiation splits Cl2 into two Cl• radicals under UV light; propagation step one has Cl• abstracting an H from CH4 to give CH3• and HCl; propagation step two has CH3• reacting with Cl2 to reform CH3Cl and a fresh Cl• radical; termination is any two radicals combining.

The single most common error is using full double-barbed curly arrows for homolytic fission instead of single-barbed ones — an instant lost mark under IB markschemes. Other frequent losses: forgetting UV light as a stated condition, missing a valid termination step, or writing the radical as CH3+ instead of CH3•.

Both levels study the reaction and its overall equation, but only HL students must write the full three-step mechanism with curly arrows. SL students need to explain that UV light causes homolytic fission and that a mixture of products forms, without drawing every propagation step in detail.

Free-radical substitution needs UV light and runs through three stages — initiation, propagation, termination — with radicals as intermediates. Nucleophilic substitution needs no light, involves a nucleophile's lone pair attacking an electrophilic carbon, and runs as a single SN1 or SN2 step. They sit in entirely separate parts of the organic chemistry syllabus.

Alkanes are saturated and largely unreactive — their C–H and C–C bonds are strong and non-polar, so there's no electrophilic or nucleophilic site to start a normal reaction. UV light supplies enough energy to homolytically break the weaker Cl–Cl or Br–Br bond, generating radicals reactive enough to attack those C–H bonds.

Alkenes carry a reactive, electron-rich C=C double bond that attracts electrophiles directly, so they add across that bond without needing UV light. Alkanes have no such site — every bond is single and non-polar — so the only way to react with a halogen is substitution via a radical mechanism triggered by light.

Radical mechanisms are usually the first time your child has to hold three separate steps in their head at once, plus unfamiliar arrow notation — it's not the chemistry that's hard, it's the working-memory load. Most students I've taught get it once they've drawn the mechanism from scratch five or six times, not just read it.

The best thing you can do isn't re-teaching the chemistry — it's handing them blank paper and asking them to draw the mechanism from memory, timed, without notes. Past paper questions on this exact topic (Reactivity 3.4 in the current DP Chemistry guide) are short and mark-heavy, so repetition beats re-reading every time.